Cantilever with combined loading — UDL over full length and an upward point load at the free end A prismatic cantilever beam of length L carries a uniformly distributed load of total magnitude W acting over its entire length. In addition, an upward concentrated force of magnitude W acts at the free end. Considering linear elastic behaviour, what will be the net deflection at the free end (direction)?

Difficulty: Medium

Correct Answer: upward

Explanation:


Introduction / Context:
This problem tests superposition for beam deflections. A cantilever with both a uniformly distributed load (UDL) and a point load at the tip can be treated as two separate load cases. The algebraic sum of the individual tip deflections gives the net deflection and its direction at the free end.


Given Data / Assumptions:

  • Cantilever beam of length L, constant E and I.
  • Uniformly distributed load with total magnitude W over the full length (so intensity w = W/L).
  • Upward point load P = W acting at the free end.
  • Small deflection Euler–Bernoulli beam theory and linear superposition are valid.


Concept / Approach:

Use standard tip-deflection formulae and add effects with signs. For a cantilever with a tip point load P (acting upward), δ_tip = + PL^3 / (3EI). For a cantilever with a UDL of intensity w (downward), δ_udl = − wL^4 / (8EI). Here w = W/L, so δ_udl = − WL^3 / (8EI). The net tip deflection is their algebraic sum.


Step-by-Step Solution:

Given P = W and w = W/L.δ_point = + WL^3 /(3EI).δ_udl = − (W/L)L^4 /(8EI) = − WL^3 /(8EI).δ_net = δ_point + δ_udl = WL^3 /(EI) * (1/3 − 1/8) = WL^3 /(EI) * (5/24).Since 5/24 > 0, deflection is upward.


Verification / Alternative check:

If instead W meant UDL intensity (force/length), the two magnitudes would not be directly comparable. The wording explicitly says “total magnitude W over its entire length”, so the above interpretation is consistent and yields a clear direction (upward).


Why Other Options Are Wrong:

  • Downward: would require the UDL effect to exceed the tip-load effect; numerically it does not for equal W as defined.
  • Zero: cancellation would require W values chosen to satisfy WL^3 /(3EI) = WL^3 /(8EI), which is not true.
  • None of these: not applicable because a unique direction (upward) is obtained.


Common Pitfalls:

  • Confusing W as total load with w as load intensity; always check units.
  • Forgetting the sign convention when superposing deflections.


Final Answer:

upward.

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